Graph y=x^2-2|x|-3

$y = x^{2} - 2 | x | - 3$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = x^{2} - 2 | x | - 3$ is $(0, - 3)$ .

To find the $x$ coordinate of the vertex, set the inside of the absolute value $x$ equal to $0$ . In this case, $x = 0$ .

$x = 0$

Replace the variable $x$ with $0$ in the expression.

$y = {(0)}^{2} - 2 | 0 | - 3$

Simplify ${(0)}^{2} - 2 | 0 | - 3$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 0 - 2 | 0 | - 3$

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$y = 0 - 2 \cdot 0 - 3$

Multiply $- 2$ by $0$ .

$y = 0 + 0 - 3$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 - 3$

Subtract $3$ from $0$ .

$y = - 3$

The absolute value vertex is $(0, - 3)$ .

$(0, - 3)$

Step 2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

For each $x$ value, there is one $y$ value. Select a few $x$ values from the domain. It would be more useful to select the values so that they are around the $x$ value of the absolute value vertex.

Substitute the $x$ value $- 2$ into $f (x) = x^{2} - 2 | x | - 3$ . In this case, the point is $(- 2, - 3)$ .

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = {(- 2)}^{2} - 2 | - 2 | - 3$

Simplify the result.

Simplify each term.

Raise $- 2$ to the power of $2$ .

$f (- 2) = 4 - 2 | - 2 | - 3$

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$f (- 2) = 4 - 2 \cdot 2 - 3$

Multiply $- 2$ by $2$ .

$f (- 2) = 4 - 4 - 3$

Simplify by subtracting numbers.

Subtract $4$ from $4$ .

$f (- 2) = 0 - 3$

Subtract $3$ from $0$ .

$f (- 2) = - 3$

The final answer is $- 3$ .

$y = - 3$

Substitute the $x$ value $- 1$ into $f (x) = x^{2} - 2 | x | - 3$ . In this case, the point is $(- 1, - 4)$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = {(- 1)}^{2} - 2 | - 1 | - 3$

Simplify the result.

Simplify each term.

Raise $- 1$ to the power of $2$ .

$f (- 1) = 1 - 2 | - 1 | - 3$

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$f (- 1) = 1 - 2 \cdot 1 - 3$

Multiply $- 2$ by $1$ .

$f (- 1) = 1 - 2 - 3$

Simplify by subtracting numbers.

Subtract $2$ from $1$ .

$f (- 1) = - 1 - 3$

Subtract $3$ from $- 1$ .

$f (- 1) = - 4$

The final answer is $- 4$ .

$y = - 4$

Substitute the $x$ value $2$ into $f (x) = x^{2} - 2 | x | - 3$ . In this case, the point is $(2, - 3)$ .

Replace the variable $x$ with $2$ in the expression.

$f (2) = {(2)}^{2} - 2 | 2 | - 3$

Simplify the result.

Simplify each term.

Raise $2$ to the power of $2$ .

$f (2) = 4 - 2 | 2 | - 3$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$f (2) = 4 - 2 \cdot 2 - 3$

Multiply $- 2$ by $2$ .

$f (2) = 4 - 4 - 3$

Simplify by subtracting numbers.

Subtract $4$ from $4$ .

$f (2) = 0 - 3$

Subtract $3$ from $0$ .

$f (2) = - 3$

The final answer is $- 3$ .

$y = - 3$

The absolute value can be graphed using the points around the vertex $(0, - 3), (- 2, - 3), (- 1, - 4), (1, - 4), (2, - 3)$

$\begin{array}{c} x & y \\ - 2 & - 3 \\ - 1 & - 4 \\ 0 & - 3 \\ 1 & - 4 \\ 2 & - 3 \end{array}$

Step 4

Graph y=x^2-2|x|-3

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